We know that ABC is an isosceles triangle, therefore angle ACB = 80 as:
180 - 20 = 160
160/2 = 80
Now, the next few parts, you need to draw on paper to understand:
First, draw an isosceles triangle with the base of it on AD, and label the angles as (80, 80 and 20) on the right hand side. Also, label the top vertex of this triangle as X.
Then on the left hand side, draw an equilateral triangle with the base as AB, with the top vertex pointing to the top left, and label all the angles as 60 as it is equilateral. Also, label the top of this vertex as Y.
Now you should be able to see a large 5-sided shape with the vertexes of YAXDB.
We can label angle CDB as 80 because triangle BDC is an isosceles, therefore the bases are the same.
We can work out angle ADB as angles in a straight line add up to 180: 180 - 80 = 100.
Label angle ADB as 100
We can find the sum of the interior angles of a 5 sided shape with: (n - 2) x 180 which if we input n = 5, we get 540.
Add up all the angles in the 5-sided shape, which will leave out ABD (from angle BYA clockwise):
60 + 60 + 20 + 80 + 20 + 80 + 100 + 60 = 480
Now we just need to work out what angle ABD is by subtracting 480 from the sum of interior angles (540) which gets: 60
From earlier on, we know angle ABC = 80 as the ABC triangle is isosceles, to work out what angle DBC is, we just need to do the final calculation:
The preimeter is adding all sides together. I took 154-55 twice sense we have four sides of a rectangle and two are lengths. after that i got 44. I then just remembered we had two sides left so divided that by 2 and got 22
